How to define a linear order on the set of $m$-ary operations on an $n$-element set?

Question

Let $m$ and $n$ be positive integers, let $S$ be a linearly ordered $n$-element set, and consider the set of all $m$-ary operations on $S$. What is the most "natural" way to linearly order the set of all $m$-ary operations on $S$, given a linear order on $S$? I am most interested in the case $m=n=2$, where $S=\{0,1\}$, linearly ordered like so: $0<1$.

Answer 1

I’d say it would be lexicographic. So if the linear order on $S$ is $$s_1<s_2<\cdots<s_n,$$ then consider for every $m$-ary operation $\star$ the $n^m$-tuple of its values on all of $S^m$. Thus for $n=m=2$, the 4-tuple for an operation $\star$ would be would be $$(\star(0,0), \star(0,1), \star(1,0), \star(1,1)),$$ where the components are arranged lexicographically by their operands. Then for any operations $\star_1$ and $\star_2$, order them lexicographically by the “coordinates” of their 4-tuples. Thus, $$\star_1 < \star_2 \mbox{ precisely when }\\ 0 \star_1 0 < 0 \star_2 0 \mbox{ or } \\ 0 \star_1 0 = 0 \star_2 0 \mbox{ and } 0 \star_1 1 < 0 \star_2 1 \mbox{ or } \\ \ldots $$